I have the following problems understanding the above circuit:

• I understand the part where the optocoupler activates transistor T3 acting as a relay driver. I also understand that this would activate relay RL1. But how does relay RL(a) get activated to N/O?
  
• Also, when the relay is in the N/C position, how does power from the 12V SMF battery get to the output?
  
• The NAND gate N1 has both inputs at the same value. When the ac mains is working, wouldn't the NAND give 0 as output where as it should actually be giving 1?

Note: This circuit can be found at http://bestengineeringprojects.com/electronics-projects/advance-mini-ups/ along with an explanation. I can not make sense of the explainaion, hence this post.

 

hi,
Look at this marked up diagram, the Red arrows and text are for mains ON.
E

 

The labeled diagram is quite helpful. Thanks Eric!
Could you confirm if its safe to conclude that once RL1 gets energized, all 4 of its contact sets switch to N/O.

 

Could you confirm if its safe to conclude that once RL1 gets energized, all 4 of its contact sets switch to N/O.

hi giv,
Yes, the relay has 4 pole change over, sets of contacts.
N/O = Normally Open, N/C = Normally Closed, this is when the relay is not energised == Off.

So all the N/O contacts are closed, when the relay is energised.

E